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# Continuity ( Section 1.8)

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Continuity ( Section 1.8). Alex Karassev. Definition. A function f is continuous at a number a if Thus, we can use direct substitution to compute the limit of function that is continuous at a. Some remarks. Definition of continuity requires three things: - PowerPoint PPT Presentation
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• Continuity (Section 1.8) Alex Karassev

• DefinitionA function f is continuous at a number a if

Thus, we can use direct substitution to compute the limit of function that is continuous at a

• Some remarksDefinition of continuity requires three things:f(a) is defined (i.e. a is in the domain of f) existsLimit is equal to the value of the functionThe graph of a continuous functions does not have any "gaps" or "jumps"

• Continuous functions and limitsTheorem Suppose that f is continuous at b andThen

Example

• Properties of continuous functionsSuppose f and g are both continuous at aThen f + g, f g, fg are continuous at aIf, in addition, g(a) 0 then f/g is also continuous at aSuppose that g is continuous at a and f is continuous at g(a). Then f(g(x)) is continuous at a.

• Which functions are continuous?TheoremPolynomials, rational functions, root functions, power functions, trigonometric functions, exponential functions, logarithmic functions are continuous on their domainsAll functions that can be obtained from the functions listed above using addition, subtraction, multiplication, division, and composition, are also continuous on their domains

• ExampleDetermine, where is the following function continuous:

• SolutionAccording to the previous theorem, we need to find domain of fConditions on x: x 1 0 and 2 x >0Therefore x 1 and 2 > xSo 1 x < 2Thus f is continuous on [1,2)

• Intermediate Value Theorem

• DefinitionsA solution of equation is also called a root of equationA number c such that f(c)=0 is called a root of function f

• Intermediate Value Theorem (IVT)f is continuous on [a,b]N is a number between f(a) and f(b)i.e f(a) N f(b) or f(b) N f(a)then there exists at least one c in [a,b] s.t. f(c) = Nxyay = f(x)f(a)f(b)bNc

• Intermediate Value Theorem (IVT)f is continuous on [a,b]N is a number between f(a) and f(b)i.e f(a) N f(b) or f(b) N f(a)then there exists at least one c in [a,b] s.t. f(c) = Nxyay = f(x)f(a)f(b)bNc1c2c3

• Equivalent statement of IVTf is continuous on [a,b]N is a number between f(a) and f(b), i.e f(a) N f(b) or f(b) N f(a)then f(a) N N N f(b) N or f(b) N N N f(a) N so f(a) N 0 f(b) N or f(b) N 0 f(a) NInstead of f(x) we can consider g(x) = f(x) Nso g(a) 0 g(b) or g(b) 0 g(a)There exists at least one c in [a,b] such that g(c) = 0

• Equivalent statement of IVTf is continuous on [a,b]f(a) and f(b) have opposite signsi.e f(a) 0 f(b) or f(b) 0 f(a)then there exists at least one c in [a,b] s.t. f(c) = 0xyay = f(x)f(a)f(b)bN = 0c

• Continuity is important!Let f(x) = 1/xLet a = -1 and b = 1f(-1) = -1, f(1) = 1However, there is no c such that f(c) = 1/c =0

• Important remarksIVT can be used to prove existence of a root of equationIt cannot be used to find exact value of the root!

• Example 1Prove that equation x = 3 x5 has a solution (root)RemarksDo not try to solve the equation! (it is impossible to find exact solution)Use IVT to prove that solution exists

• Steps to prove that x = 3 x5 has a solution Write equation in the form f(x) = 0x5 + x 3 = 0 so f(x) = x5 + x 3 Check that the condition of IVT is satisfied, i.e. that f(x) is continuousf(x) = x5 + x 3 is a polynomial, so it is continuous on (-, )Find a and b such that f(a) and f(b) are of opposite signs, i.e. show that f(x) changes sign (hint: try some integers or some numbers at which it is easy to compute f)Try a=0: f(0) = 05 + 0 3 = -3 < 0Now we need to find b such that f(b) >0Try b=1: f(1) = 15 + 1 3 = -1 < 0 does not workTry b=2: f(2) = 25 + 2 3 =31 >0 works!Use IVT to show that root exists in [a,b]So a = 0, b = 2, f(0) 0 and therefore there exists c in [0,2] such that f(c)=0, which means that the equation has a solution

• x = 3 x5 x5 + x 3 = 0xy0-3312N = 0c (root)

• Example 2Find approximate solution of the equation x = 3 x5

• Idea: method of bisectionsUse the IVT to find an interval [a,b] that contains a rootFind the midpoint of an interval that contains root: midpoint = m = (a+b)/2 Compute the value of the function in the midpoint If f(a) and f (m) are of opposite signs, switch to [a,m] (since it contains root by the IVT), otherwise switch to [m,b]Repeat the procedure until the length of interval is sufficiently small

• f(x) = x5 + x 3 = 002f(x)x-331We already know that [0,2] contains rootMidpoint = (0+2)/2 = 1-1< 0> 0

• f(x) = x5 + x 3 = 002f(x)x-3311-11.56.1Midpoint = (1+2)/2 = 1.5

• f(x) = x5 + x 3 = 002f(x)x-3311-11.56.1Midpoint = (1+1.5)/2 = 1.251.251.3

• 1-11.251.31.125-.07f(x) = x5 + x 3 = 002f(x)x-3311.56.1Midpoint = (1 + 1.25)/2 = 1.125By the IVT, interval [1.125, 1.25] contains rootLength of the interval: 1.25 1.125 = 0.125 = 2 / 16 = = the length of the original interval / 2424 appears since we divided 4 timesBoth 1.25 and 1.125 are within 0.125 from the root!Since f(1.125) -.07, choose c 1.125Computer gives c 1.13299617282...

• ExerciseProve that the equation sin x = 1 x2 has at least two solutionsHint:Write the equation in the form f(x) = 0 and find three numbers x1, x2, x3,such that f(x1) and f(x2) have opposite signs AND f(x2) and f(x3) have opposite signs. Then by the IVT the interval [ x1, x2 ] contains a root AND the interval [ x2, x3 ] contains a root.

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